Question 50280: Find three consecutive even integers such that twice the sum of the first and third is twelve more than twice the second.
My attempt:
#1 = x #2 = x+2 #3 = x+4
2(x+x+4) - 2(x+2) =12
2x+2x+8 - 2x-4 = 12
2x-4 = 12
2x = 16
x = 8
x+2 = 10
x+4 = 12
The problem doesn't check so it's evidentally wrong, I'm thinking I set the problem up incorrectly. Can you please help me find my mistake or redirect me?
Thanks for your help.
Answer by tutorcecilia(2152)   (Show Source):
You can put this solution on YOUR website! Find three consecutive even integers such that twice the sum of the first and third is twelve more than twice the second.
Three consecutive even intergers:
First integer=x
Second integer = x+2
Third integer = x+4
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Translation:
Twice the sum of the first and third: 2[(x)+(x+4)]
The "sum" means to add the first and third integer, than multiply the total by "2".
.
"is" means "equals"
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Twelve more than twice the second: [(2)(x+2)]+12
Take the second integer (x+2), multiply it by "2" and add 12.
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So, putting it all together:
2[(x)+(x+4)]=[(2)(x+2)]+12
2(2x+4)=(2x+4)+12
4x+8=2x+16
4x-2x+8=2x-2x+16
2x+8-8=16-8
2x=8
2x/2=8/2
x=4
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Plug (x=4) into the integers and solve.
First integer=x=4
Second integer = x+2=4+2=6
Third integer = x+4=4+4=8
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2[(x)+(x+4)]=[(2)(x+2)]+12
2(4+8)=(2)(6)+12
2(12)=12+12
24=24 [checks out]
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